B

Step 1: Differentiate the function \( y = 3t^2 - 4t \) with respect to \( t \).
\[ \frac{dy}{dt} = 6t - 4
Step 2: Set the derivative equal to zero to find critical points:
\[ 6t - 4 = 0 \]
\[ 6t = 4 \]
\[ t = \frac{2}{3}
Step 3: To determine if this point is a minimum, we can use the second derivative test.
Step 4: Differentiate again: \( \frac{d^2y}{dt^2} = 6 \) which is positive.
Therefore, \( t = \frac{2}{3} \) is a minimum point.
Step 5: Find \( y \) at this point:
\[ y = 3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) \]
\[ = 3 \times \frac{4}{9} - \frac{8}{3}
= \frac{4}{3} - \frac{8}{3} \]
\[ = -\frac{4}{3}
Therefore, the minimum value of \( y \) is -\frac{4}{3}.
Hence, the correct choice is B.